A clock reaction can be used to measure the rate of reaction in certain circumstances. In a typical reaction the initial rate (at t=0) doesn’t actually change over a short period of time whilst there are still plenty of reactants present in the reaction mixture.
For most reactions this equates to the first 15% of the reaction taking place, after which the rate of reaction slows.
If we measure the time taken for the reaction to use up a certain amount of a reactant or produce a certain amount of a product, the shorter the time, the faster the reaction has occurred. This means we can take 1/t as a relative measure of the initial rate of reaction.
The trick, of course, is knowing when the fixed amount of product has been formed.
Bromate(V), bromide and hydrogen ions react according to the following equation:
5Br–(aq) + BrO3–(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l)
If we add a small amount of phenol solution and methyl orange solution to the reaction mixture, the molecular bromine produced in the main reaction reacts instantly with the phenol to form 2,4,6-tribromophenol:
As soon as the main reaction has produced sufficient bromine to react with the fixed amount of phenol, free bromine will appear in the solution. This bromine will immediately bleach the methyl orange solution:
Br2 + methyl orange (pink) → bleached methyl orange (colourless) + 2Br–
The disappearance of the pink colour tells us that a fixed amount of product has been formed by the main reaction. If we time how long it takes for this to happen, then 1/t is a direct measure of the initial rate of reaction. Essentially, we have a bromine clock reaction.
We can find the order of reaction with respect to each of the reactants by planning a series of experiments in which we vary the starting concentrations of bromide ions, bromate ions and hydrogen ions in the reaction mixture and time how long it takes for the colour to change in each experiment.
A specimen set of results for varying the starting concentration of bromide ions (and keeping the concentration of everything else constant) is shown below:
| [Br–] / mol dm-3 | time / sec | 1/t |
|---|---|---|
| 2.50 x 10-3 | 22.7 | 0.044 |
| 2.00 x 10-3 | 26.0 | 0.038 |
| 1.50 x 10-3 | 34.0 | 0.029 |
| 1.25 x 10-3 | 37.5 | 0.026 |
| 1.00 x 10-3 | 48.0 | 0.021 |
Since 1/t is a measure of the initial rate, all we need do now is draw a graph of [Br–] against 1/t to find the order of reaction with respect to the bromide ions (first order – proportional relationship).
In practice we would then repeat the experiment, firstly varying the concentration of bromate ions and secondly varying the concentration of hydrogen ions, draw graphs for each and hey presto, we know the order of reaction for these two reactants as well!
Another common example of a clock reaction is the iodine clock, of which there are several iterations, although probably the most famous of these is the reaction involving hydrogen peroxide and iodide ions in acid solution:
H2O2(aq) + 2I–(aq) + 2H+(aq) → I2(aq) + 2H2O(l)
The kinetics of this reaction were first investigated by Vernon Harcourt and William Essen, and the reaction is still referred to as the Harcourt-Essen reaction. A fixed amount of sodium thiosulphate solution together with a little starch solution is added to the reactants in each experiment. The molecular iodine produced by the main reaction between hydrogen peroxide, iodide and acid immediately reacts with the thiosulphate ions:
I2(aq) + 2S2O32– (aq) → S4O62–(aq) + 2I–(aq)
When all of the thiosulphate has been used up, the iodine accumulates in the solution and reacts with the starch to give a distinctive blue-black colour. The time from mixing the reactants to the appearance of the blue colour is therefore the time for a fixed amount of iodine (product) to be formed and we can simply use 1/t as a measure of the initial rate of reaction.
Once again, we can find the order of reaction with respect to each of the reactants by planning a series of experiments in which we vary the starting concentrations of iodide ions, hydrogen peroxide and hydrogen ions in the reaction mixture and time how long it takes for the blue-black starch complex to form in each experiment.
You can watch a good video of the practical here if you haven’t done this experiment yourself 😊.
How do we know how much of the reaction is being studied?
Let’s take a look at another version of the iodine clock – the reaction between peroxodisulphate ions and iodide ions.
S2O82-(aq) + 2I–(aq) ⇾ 2SO42-(aq) + I2(aq)
We can follow the reaction by following the colour of iodine as its is produced which is much more obvious if we add little starch to our reaction mixture. As iodine is produced, it forms a starch-iodine complex which is a distinctive blue-black colour.
We can measure how long it takes to produce a small, fixed amount of iodine by adding thiosulphate ions, just as we did in the Harcourt-Essen reaction.
I2(aq) + 2S2O32– (aq) → S4O62–(aq) + 2I–(aq)
The thiosulphate ions reduce molecular iodine back to iodide ions until all the thiosulphate is used up, at which point the blue-black iodine-starch complex appears instantly. We have our ‘clock’.
The method for the experiment is straightforward. On addition of the potassium peroxodisulphate to the other reactants, the stop watch is started and stopped when the blue-black colour appears. The time taken is recorded, and 1/t (the initial rate of reaction) calculated.
Note that in the method above, the actual concentration of iodide ions in the reaction mixture is
(volume of KI added / total volume of reaction mixture) x concentration of KI = (5/10) x 1.0 = 0.5 mol dm-3
Now let’s work out the extent of the reaction being studied as a percentage (remember that it should be less than 15% for a clock reaction to be accurate enough to assume we are studying the initial rate).
- We need to work out which issue the limiting reactant in the main reaction, so we can determine how much iodine the reaction will make if it runs to completion.
S2O82-(aq) + 2I–(aq) ⇾ 2SO42-(aq) + I2(aq)
- no. of mol of S2O82- = concentration x volume (dm3) = 0.04 x 0.002 = 8.0 x 10-5 mol
ratio S2O82- : I2 is 1:1 so maximum amount of I2 possible is 8.0 x 10-5 mol
- no. of mol of I– = concentration x volume (dm3) = 1.0 x 0.005 = 5.0 x 10-3 mol
ratio I– : I2 is 2:1 so maximum amount of I2 possible is 2.5 x 10-3 mol
Therefore potassium peroxodisulphate is the limiting reactant and the maximum amount of iodine that the reaction can make is 8.0 x 10-5 mol.
- Now we need to determine the amount of sodium thiosulphate present because this will give us the amount ( number of moles) of iodine used up in the ‘clock’.
I2(aq) + 2S2O32– (aq) → S4O62–(aq) + 2I–(aq)
- no. of mol of S2O32- = concentration x volume (dm3) = 0.01 x 0.002 = 2.0 x 10-5 mol
- ratio S2O32– : I2 is 2:1 so the number of moles of I2 that react in before the blue-black complex forms is 1.0 x 10-5 mol.
- % of reaction studied = (1.0 x 10-5 / 8.0 x 10-5) x 100 = 12.5%